A truck starts from rest and accelerates uniformly at 2.0 m s^{–2}. At *t *= 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at *t *= 11 s? (Neglect air resistance.)

#### Solution 1

**(a)** 22.36 m/s, at an angle of 26.57° with the motion of the truck

**(b)** 10 m/s^{2}

**(a) **Initial velocity of the truck, *u* = 0

Acceleration, *a* = 2 m/s^{2}

Time, *t* = 10 s

As per the first equation of motion, final velocity is given as:

*v = u + at*

= 0 + 2 × 10 = 20 m/s

The final velocity of the truck and hence, of the stone is 20 m/s.

At *t* = 11 s, the horizontal component (*v*_{x}) of velocity, in the absence of air resistance, remains unchanged, i.e.,

*v*_{x} = 20 m/s

The vertical component (*v*_{y}) of velocity of the stone is given by the first equation of motion as:

*v*_{y}* = u + a*_{y}δ*t*

Where, δ*t* = 11 – 10 = 1 s and* a*_{y} = g = 10 m/s^{2}

∴*v*_{y} = 0 + 10 × 1 = 10 m/s

The resultant velocity (*v*) of the stone is given as:

v = `sqrt(v_x^2+v_y^2)`

`= sqrt(20^2 + 10^2) = sqrt(400 + 100)`

=`sqrt(500)` = 22.36 m/s

Let *θ* be the angle made by the resultant velocity with the horizontal component of velocity, *v*_{x}

_{`:. tan theta = (v_y)/(v_x)`}

`theta = tan^(-1)(10)/(20)`

`=tan^(-1)(0.5)`

= `26.57^@`

**(b) **When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s^{2} and it acts vertically downward.

#### Solution 2

u = 0, a = 2 ms^{-2}, t 10 s

Using equation, v = u + at, we get

v = 0 + 2 x 10 = 20 ms^{-1}

(a) Let us first consider horizontal motion. The only force acting on the stone is force of gravity which acts vertically downwards.

Its horizontal component is zero. Moreover, air resistance is to be neglected. So, horizontal motion is uniform motion.

∴ vx = v = 20 ms^{-1}

Let us now consider vertical motion which is controlled by force of gravity.

u=0, a = g = 10 ms^{-2}, t = (11 — 10) s = 1 s

Using `v = u + at, v_y = 0 + 10 xx 1 = 10 m/s^(-1)`

Resultant velocity,

`v = sqrt(v_x^2 + v_y^2)`

=>v = `sqrt(20^2 + 10^2) ms^(-1)`

`=sqrt(500) ms^(-1)`

= 22.36 `ms^(-1)`

`tan beta = v_y/v_x = 10/20 = 1/2 = 0.5`

or `beta = tan^(-1) (0.5) = 26.56^@`

or β = 26º 34'. This angle is with the horizontal

(b) The moment the stone is dropped from the car, horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity. This gives a vertically downward acceleration of 10 ms-2. This is also the net acceleration of the stone.